1. Calculating specific heat capacities

2. A V Calculating the specific heat capacity of aluminium

3. 3 600.00 Aluminium block mass 0.5kg Current = 1.5A Pd across heater = 6V Total electrical energy in E= IVΔt = 1.5A x 6V x 3600s = 32 400J Initial temperature = 20 C Final temperature = 85C

4. Total electrical energy in 32 400J is taken as equal to Q the heat energy transferred to the block Q= mcΔθ So c = Q/mΔθ c = 32 400/ (0.5kg x 65K) =997Jkg -1 K -1 The actual specific heat capacity of Al is 900 Jkg -1 K -1 The discrepancy is due to heat loss. We can reduce heat losses in the normal way by insulating the system and use glycerine around the thermometer to improve thermal contact A rather clever way to compensate is to “correct for cooling” using a cooling correction factor which can be determined experimentally (this method is beyond the scope of the course)

5. A V Liquid under test stirrer Still air reduces heat loss by conduction and convection

6. SubstancePhaseSpecific heat capacity J·g -1 ·K -1 Hydrogengas14.300 Ironsolid0.444 Lithiumsolid3.582 Mercuryliquid0.139 Nitrogengas1.042 Oilliquid≈ 2.000 Oxygengas0.920 SilicaSilica (fused)solid0.703 Watergas2.020 liquid4.183 solid (0 °C)2.060

7. Table of shc SubstancePhaseSpecific heat capacity J·g -1 ·K -1 AirAir (dry)gas1.005 AirAir (100% humidity)gas≈ 1.030 Aluminiumsolid0.900 Berylliumsolid1.824 Brasssolid0.377 Coppersolid0.385 Diamondsolid0.502 Ethanolliquid2.460 Goldsolid0.129 Graphitesolid0.720 Heliumgas5.190

8. An aluminium saucepan of mass 1.2kg contains 0.5 kg of pure water. It is heated and the temperature of the pan and contents rise together as show in the graph below. Given the specific heat capacities of aluminium and water find the rate at which energy is supplied to both the pan and water.

9. Aluminium has a specific heat capacity of 900Jkg -1 K -1 Water has a specific heat capacity of 4 200Jkg -1 K -1

10. This aluminium saucepan has a mass of 1.2 kg Its heat capacity is 1.2 x 900 = 1080 JK -1 The water in it has a mass of 0.5kg The heat capacity of this mass of water is 0.5 x 4200 = 2100JK -1 So the total heat capacity of the aluminium and the water is 1080 + 2100 = 3180JK -1

11. Find the rate of temperature rise at the beginning of the heating process i.e. before too much evaporation of the liquid takes place which brings other factors into play. 10 min 30 C Degrees K per second

12. For the pan and water together Heat capacity = 3180JK -1 Temperature rise = 0.05Ks -1 P.S. If you are doing this you should get around 230 W as the gradient of the graph that I have drawn is slightly different from that given in the question Rate of energy change = Heat capacity x rate of temperature rise